We are given two functions: $f$ and $f \circ g$ in the form of ordered pairs. To determine $g$, we can use the relation $f \circ g(x) = f(g(x))$, which gives us a clue for calculating the corresponding values of $g$.

Problem breakdown:

• $f = \{(3, -4), (1, 8), (2, -5), (0, 4), (10, 1)\}$
• $f \circ g = \{(2, 8), (1, -5), (0, -4), (3, 4)\}$

We need to:

1. Determine the function $g$.
2. Verify the composition $g \circ f$.

Step-by-step solution:

Part a: Finding $g$

Since $f \circ g(x) = f(g(x))$, we know that the output of $f(g(x))$ must match the second element of each pair in $f \circ g$. Let's find $g(x)$ for each pair:

1. From $f \circ g(2) = 8$, we need $f(g(2)) = 8$. In the function $f$, $f(1) = 8$. Therefore, $g(2) = 1$.
2. From $f \circ g(1) = -5$, we need $f(g(1)) = -5$. In the function $f$, $f(2) = -5$. Therefore, $g(1) = 2$.
3. From $f \circ g(0) = -4$, we need $f(g(0)) = -4$. In the function $f$, $f(3) = -4$. Therefore, $g(0) = 3$.
4. From $f \circ g(3) = 4$, we need $f(g(3)) = 4$. In the function $f$, $f(0) = 4$. Therefore, $g(3) = 0$.

Thus, the function $g$ is: $g = \{(2, 1), (1, 2), (0, 3), (3, 0)\}$

Part b: Finding $g \circ f$

To find $g \circ f(x) = g(f(x))$, we apply the function $g$ to the output of $f(x)$:

1. $f(3) = -4$, but $g(-4)$ is undefined, so this pair does not contribute.
2. $f(1) = 8$, but $g(8)$ is undefined, so this pair does not contribute.
3. $f(2) = -5$, but $g(-5)$ is undefined, so this pair does not contribute.
4. $f(0) = 4$, but $g(4)$ is undefined, so this pair does not contribute.
5. $f(10) = 1$, and $g(1) = 2$. Therefore, $g \circ f(10) = 2$.

Thus, the only valid pair for $g \circ f$ is: $g \circ f = \{(10, 2)\}$

Summary:

• $g = \{(2, 1), (1, 2), (0, 3), (3, 0)\}$
• $g \circ f = \{(10, 2)\}$

Let me know if you'd like any further details or explanations!

Here are some related questions for further exploration:

1. How do you generalize the process of finding the inverse of a function composition?
2. What happens when a function is not injective or surjective? How does it affect compositions?
3. Can we always find $g$ given $f \circ g$ and $f$? What if some values of $f$ are undefined?
4. What are the conditions for two functions to be composable in terms of their domain and codomain?
5. How can the concept of function composition be extended to multiple functions, e.g., $f \circ g \circ h$?

Tip: Function composition is only possible if the output of the second function is a valid input for the first one, so always check their domains carefully!